The discussion revolves around solving the trigonometric equation $\sin a \cos b + \sin b \cos c + \sin c \cos a = \frac{3}{2}$ using the Cauchy–Schwarz inequality and the AM-GM inequality. Participants explore the implications of these inequalities in the context of the equation, examining potential solutions and the conditions under which equality holds.
There appears to be a general agreement on the application of the Cauchy–Schwarz and AM-GM inequalities, as well as the resulting conditions for the angles. However, the discussion does not resolve whether there are additional solutions or alternative methods to approach the problem.
The discussion relies on specific assumptions about the angles and the properties of the sine and cosine functions. The implications of the inequalities are contingent on these assumptions, and the exploration of solutions is limited to the specified interval.
Ackbach said:$a=b=c=\dfrac{ \pi}{4}$ solves the equation by inspection. There may be other solutions.
Ackbach said:Fine. Be that way! How about $a,b,c \in \left\{ \dfrac{ \pi}{4} \pm n \pi \bigg| n \in \mathbb{Z} \right \}?$
Opalg said:[sp]Using first the Cauchy–Schwarz inequality and then the AM-GM inequality, $$\begin{aligned}\sin a \cos b+ \sin b \cos c+ \sin c \cos a &\leqslant \sqrt{(\sin^2a + \sin^2 b + \sin^2c)(\cos^2a + \cos^2b + \cos^2c)} \\ &\leqslant \tfrac12(\sin^2a + \sin^2 b + \sin^2c + \cos^2a + \cos^2b + \cos^2c) = \tfrac32.\end{aligned}$$ Equality in the Cauchy–Schwarz inequality only occurs if the vectors are scalar multiples of each other. In this case, that means that there exists a scalar $\lambda$ such that $$\sin a = \lambda\cos b, \qquad \sin b = \lambda\cos c, \qquad \sin c = \lambda\cos a.$$ Equality in the AM-GM inequality only occurs if the numbers are equal. In this case, that means that $\lambda^2(\cos^2a + \cos^2b + \cos^2c) = (\cos^2a + \cos^2b + \cos^2c)$, so $\lambda^2 = 1$. It then follows that $$\sin^2a = \cos^2b = 1-\sin^2b = 1-\cos^2c = \sin^2c = \cos^2a.$$ Thus $\cos a = \pm\sin a$ and similarly for $b$ and $c$. So each of $a$, $b$, $c$ is an odd multiple of $\pi/4$. Each of the three products $\sin a \cos b$, $\sin b \cos c$ and $\sin c \cos a$ must be $\pm1/2$, and in fact they must all be $+1/2$ if their sum is to be $3/2$. So $\cos b$ must have the same sign as $\sin a$, and similarly for the other two products. There are thus eight solutions in any interval of length $2\pi$, given by the possible choices of sign in the equations $$\sin a = \cos b = \pm\tfrac1{\sqrt2},\qquad \sin b = \cos c = \pm\tfrac1{\sqrt2},\qquad \sin c = \cos a = \pm\tfrac1{\sqrt2}.$$ In the interval $[-\pi,\pi]$ the solutions are: $$\begin{array}{c|c|c} a&b&c \\ \hline \pi/4 & \pi/4 & \pi/4 \\ 3\pi/4 & \pi/4 & -\pi/4 \\ \pi/4 & -\pi/4 & 3\pi/4 \\ 3\pi/4 & -\pi/4 & -3\pi/4 \\ -\pi/4 & 3\pi/4 & \pi/4 \\ -\pi/4 & -3\pi/4 & 3\pi/4 \\ -3\pi/4 & 3\pi/4 & -\pi/4 \\ -3\pi/4 & -3\pi/4 & -3\pi/4 \\ \end{array}$$[/sp]